Re: design example
- From: "Andrew Holme" <andrew@xxxxxxxxxx>
- Date: Sun, 24 Apr 2005 10:10:55 +0100
Mauri wrote:
> Hi all,
> i need your help for understanding this design example
>
> Problem: A full bridge speed controller circuit is designed to control
> a 12v motor. The switching frequency must be above the audible limit
> (20kHz). The motor has a total resistance of 0.12 Ohms. Choose
> suitable MOSFETs for the bridge circuit, within a reasonable price
> limit, and suggest any heatsinking that may be required. The ambient
> temperature is assumed to be 25ºC.
>
> Solution: Lets have a look at the IRF3205 and see if it is suitable.
> First the drain current requirement. At stall, the motor will take 12v
> / 0.12 Ohms = 100 Amps. We will first make a guess at the junction
> temperature, at 125ºC We must find what the maximum drain current is
> at 125ºC first. The graph of figure 9 shows us that at 125ºC, the
> maximum drain current is about 65 Amps. Therefore 2 IRF3205s in
> parallel should be capable in this respect.
> How much power will the two parallel MOSFETs be dissipating? Lets
> start with the power dissipation whilst ON and the motor stalled, or
> just starting. That is the current squared times the on-resistance.
> What is RDS(on) at 125ºC? Figure 4 shows how it is derated from its
> front-page value of 0.008 Ohms, by a factor of about 1.6. Therefore,
> we assume RDS(on) will be 0.008 x 1.6 = 0.0128. Therefore PD = 50 x 50
> x 0.0128 = 32 Watts. How much of the time will the motor be either
> stalled or starting? This is impossible to say, so we will have to
> guess. 20% of the time is quite a conservative figure - it is likely
> to be a lot less. Since the power causes heat, and the heat conduction
> is quite a slow process, the effect of power dissipation tends to get
> averaged out over quite long time periods, in the region of seconds.
> Therefore we can derate the power requirement with the quoted 20%, to
> arrive at an average power dissipation of 32W x 20% = 6.4W.
> Now we must add the power dissipated due to switching. This will occur
> during the rise and fall times, which are quoted in the Electrical
> Characteristics table as 100ns and 70ns respectively. Assuming the
> MOSFET driver can supply enough current to fulfill the requirements of
> these figures (gate drive source resistance of 2.5 Ohms = pulse output
> drive current of 12v / 2.5 Ohms = 4.8 Amps), then the ratio of
> switching time to steady-state time is 170ns * 20kHz = 3.4mW which is
I take it you mean micro-seconds there.
> negligable. These on-off timings are a bit crude however, for more
> information about on-off times, see here.
> Now what are the switching requirements? The MOSFET driver ship we use
> will cope with most of these, but its worth checking. The turn-on
> voltage, Vgs(th), from the graphs of Figure 3 is just over 5 Volts. We
> have already seen that the driver should be able to source 4.8 Amps
> for a very short period of time.
> Now what about the heatsink. You may want to read the chapter on
> heatsinks before this section. We want to keep the temperature for the
> semiconductor junction below 125ºC, and we have been told that the
> ambient temperature is 25ºC. Therefore, with a MOSFET dissipating 6.4W
> on average, the total thermal resistance must be less than (125 - 25)
> / 6.4 = 15.6 ºC/W. The thermal resistance from junction to case makes
> up for 0.75 ºC/W of this, typical case to heatsink values (using
> thermal compound) are 0.2 ºC/W, which leaves 15.6 - 0.75 - 0.2 = 14.7
> ºC/W for the heatsink itself. Heatsinks of this θjc value are
> quite small and cheap. Note that the same heatsink can be used for
> both MOSFETs to the left of or to the right of the load in the H-
> bridge, since these two MOSFETs are never both on at the same time,
> and so can never both be dissipating power at the same time. The cases
> of them must be electrically isolated however. See the heatsinks page
> for more information on the required electrical isolation.
>
> These are my doubts:
>
> 1) "Assuming the MOSFET driver can supply enough current to fulfill
> the requirements of these figures (gate drive source resistance of 2.5
> Ohms = pulse output drive current of 12v / 2.5 Ohms = 4.8 Amps)". What
> does it means ? Shouldn't the choise of the mosfet driver be based on
> Qg= I*T --> I=Qg/T --> I=146ns*20KHz = 2,92mA ?
The gate has to be charged and discharged very quickly, by short pulses
taking of the order of 100 nanoseconds or less, each. I think you are using
the wrong value for time in this calculation, but I'm a bit confused by your
use of 'ns' for units of charge. I take it this is meant to be 146
nano-Coulombs total gate charge?
.
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